# Half Angle Formulas/Cosine

## Theorem

 $\displaystyle \cos \frac \theta 2$ $=$ $\displaystyle + \sqrt {\frac {1 + \cos \theta} 2}$ for $\dfrac \theta 2$ in quadrant I or quadrant IV $\displaystyle \cos \frac \theta 2$ $=$ $\displaystyle - \sqrt {\frac {1 + \cos \theta} 2}$ for $\dfrac \theta 2$ in quadrant II or quadrant III

where $\cos$ denotes cosine.

## Proof 1

 $\displaystyle \cos \theta$ $=$ $\displaystyle 2 \cos^2 \frac \theta 2 - 1$ Double Angle Formula for Cosine: Corollary 1 $\displaystyle \leadsto \ \$ $\displaystyle 2 \cos^2 \frac \theta 2$ $=$ $\displaystyle 1 + \cos \theta$ $\displaystyle \leadsto \ \$ $\displaystyle \cos \frac \theta 2$ $=$ $\displaystyle \pm \sqrt {\frac {1 + \cos \theta} 2}$

We also have that:

In quadrant I, and quadrant IV, $\cos \dfrac \theta 2 > 0$
In quadrant II and quadrant III, $\cos \dfrac \theta 2 < 0$.

$\blacksquare$

## Proof 2

Define:

$u = \dfrac \theta 2$

Then:

 $\displaystyle \cos^2 u$ $=$ $\displaystyle \frac {1 + \cos2u} 2$ Power Reduction Formulas $\displaystyle \implies \ \$ $\displaystyle \cos \frac \theta 2$ $=$ $\displaystyle \pm \sqrt {\frac {1 + \cos \theta} 2}$

We also have that:

In quadrant I, and quadrant IV, $\cos \dfrac \theta 2 > 0$
In quadrant II and quadrant III, $\cos \dfrac \theta 2 < 0$.

$\blacksquare$